To make a 2d Minkowski space-time diagram for special relativity one typically first picks an inertial frame. The vertical axis of the graph is chosen to be the time ct axis. The horizontal axis is chosen to be a distance coordinate x. This means that velocity of a particle is obtained from the reciprocal of the slope of its path on the diagram instead of directly from the slope. Yes, this seems weird to do at first but its also the usual way to do it so you get used to it. So the path of an accelerating particle on a spacetime diagram can look like:

Now lets say the path represents the path of an accelerated observer. The events comprised of the ticks on the watch of this observer all lay along the path and so the observer's world line or the path the observer takes can be labeled as the time axis for this observer. So for special relativity the graph looks like:

(Fig. 2.1.2)

Next consider the case where the primed frame observer is in a state of constant velocity and the clocks are synchronized at the origin. In this case it becomes:

The time ct' axis consists of the events that are all at x' = 0. The Lorentz transformation equations describe this line parametrically.

The x' axis consists of events at ct' = 0. In special relativity the Lorentz transformation equations also describe this parametrically

From these we verify that the reciprocal of the ct' axis slope results in the velocity , but we also see that for the x' axis, the slope itself results in the velocity. So we can also now put the x' axis on the graph.

If a y axis is included coming out of the screen, the light paths sweep out a cone shape. In special relativity this is known as a light cone.

Exercises

Problem 2.1.1

Lines of simultaneity in S' are lines parallel to the x' axis. Superimpose such lines onto Fig 2.2.4. Are spatially displaced events along a line of simultaneity from S' also simultaneous according to the S frame?

Problem 2.1.2

Consider a tachyon (a hypothetical faster than c particle) that leaves x = 10m for x = -10m and travels the distance in almost no time according to S. Draw this onto the figure for I.2.1.1. Is it traveling forward or backward in time according to S' ? What if it traveled from x = -10 m to x = 10 m instead?

Problem 2.1.3

Plot two events on the figure that reverse in time order going from one frame to the other.

For the displacement between events in 4 dimensional space-time we should include the temporal displacement between the event in the interval for special relativity. This can be done in such a way that the displacement calculated is the same according to any inertial frame. We define the invariant interval as

(2.2.2)

which can be written

(2.2.3)

To verify that this interval has been constructed in an invariant way insert the expressions for the

differentials from the differential form of the Lorentz coordinate transformation equations Eqn 1.1.5.

After simplification, the interval reduces to the same form.

When we observe an object in motion and describe the length of its path through space-time by the invariant interval we should realize that the object does not move according to its own frame for special relativity

dx' = dy' = dz' = 0

therefor the invariant interval reduces to . Since the interval is an invariant and it is equal to the proper time of the object, we can look at proper time as an invariant.

We can choose to define a displacement four-vector with the following relations

Using Einstein summation as discussed the invariant interval is written in more compact notation as

(2.2.5)

For special relativity light paths are described by ds = 0. therefor any path given by ds = 0 is called a light-like path. A path where the overall sign of ds2 is negative is called a space-like path. A path where the

overall sign of ds2 is positive is called a time-like path.

According to the first postulate of special relativity the laws of physics are the same for every inertial frame. Therefor when modeling the general laws of physics with equations we must use equations that do not change their basic form when transformed from one frame to another.

For instance, lets say we do some tests in a lab and we find we have some natural phenomenon modeled by

F(ct,x,y,z,dxi/dxj,...) - G(ct,x,y,z,dxm/dxn,...) = 0,

If this is the general law of physics then in any other coordinate system it should also be

In element notation, using the Einstein summation convention the transformation can be written

(2.2.10)

Note -

(2.2.11)

and this transformation is just the ordinary chain rule of calculus.

For special relativity a tensor will be anything that Lorentz transforms. A contravariant tensor in special relativity will be any quantity that transforms between frames according to

(2.2.12)

A covariant tensor in special relativity will be any quantity that transforms between frames according to

(2.2.13)

There are also mixed tensors. For example

(2.2.14)

From these transformation properties we can deduce that for an individual particle,

A sum or difference of tensors is still a tensor.

A product of tensors is still a tensor.

A tensor multiplied or divided by an invariant is still a tensor.

[note - these rules apply only when the tensors involved describe that which is observed, not the state of the observer himself. So for example let be a tensor describing something observed like say the electromagnetic field and is the four-vector velocity of the observer (c,0,0,0). It turns out that the electric field given by

is NOT a tensor. As is the four-vector velocity of whoever is the observer everyone uses (c,0,0,0) as a result and the expression does not transform as a four-vector. . If were the four-vector velocity of one "particular" observer then the expression would transform as a tensor, but then it wouldn't represent the electric field to anyone except that observer and it would then only when is the electromagnetic field already expressed according to his own frame. Likewise the magnetic field

where is the four-velocity of the observer (c,0,0,0) is also not a tensor.]

In relativity we write the fundamental equations of physics as tensor equations such as

(2.2.15)

because this doesn't change its form in a frame transformation. For instance, using the above transformation properties, it is easy to show that in any other frame this equation remains in the same form

This is what satisfies the first postulate of special relativity, that the laws of physics are the same for all inertial frames.

Notice that since we model the laws of physics with tensor equations whose expressions are tensors defined by the transformation properties of the coordinates and since the Lorentz coordinate transformations are one to one invertable, there can be no true paradox's in special relativity.

Exercises

Problem 2.2.1

Use the definition of a tensor to verify that for an individual particle,

1. A sum or difference of tensors is still a tensor.

2. A product of tensors is still a tensor.

3. A tensor multiplied or divided by an invariant is still a tensor.

[note - these rules apply only when the tensors involved describe that which is observed, not the state of the observer himself. So for example let be a tensor describing something observed like say the electromagnetic field and is the four-vector velocity of the observer (c,0,0,0). It turns out that the electric field given by

is NOT a tensor. As is the four-vector velocity of whoever is the observer everyone uses (c,0,0,0) as a result and the expression does not transform as a four-vector. . If were the four-vector velocity of one "particular" observer then the expression would transform as a tensor, but then it wouldn't represent the electric field to anyone except that observer and it would then only when is the electromagnetic field already expressed according to his own frame. Likewise the magnetic field

where is the four-velocity of the observer (c,0,0,0) is also not a tensor.]